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## Integrated math 3

### Course: Integrated math 3 > Unit 4

Lesson 4: Putting it all together# Graphs of polynomials

Analyze polynomials in order to sketch their graph.

#### What you should be familiar with before taking this lesson

The $f$ describes the behavior of its graph at the "ends" of the $x$ -axis. Algebraically, end behavior is determined by the following two questions:

**end behavior**of a function- As
, what does$x\to +\mathrm{\infty}$ approach?$f(x)$ - As
, what does$x\to -\mathrm{\infty}$ approach?$f(x)$

If this is new to you, we recommend that you check out our end behavior of polynomials article.

The zeros of a function $f$ correspond to the $x$ -intercepts of its graph. If $f$ has a zero of $x$ -axis at that $x$ value. If $f$ has a zero of $x$ -axis at that point.

*odd*multiplicity, its graph will*cross*the*even*multiplicity, its graph will*touch*theIf this is new to you, we recommend that you check out our zeros of polynomials article.

#### What you will learn in this lesson

In this lesson, we will use the above features in order to analyze and sketch graphs of polynomials. We will then use the sketch to find the polynomial's positive and negative intervals.

## Analyzing polynomial functions

We will now analyze several features of the graph of the polynomial $f(x)=(3x-2)(x+2{)}^{2}$ .

### Finding the $y$ -intercept

To find the $y$ -intercept of the graph of $f$ , we can find $f(0)$ .

The $y$ -intercept of the graph of $y=f(x)$ is $(0,-8)$ .

### Finding the $x$ -intercepts

To find the $x$ -intercepts, we can solve the equation $f(x)=0$ .

$\begin{array}{rl}f(x)& =(3x-2)(x+2{)}^{2}\\ \\ {0}& =(3x-2)(x+2{)}^{2}\\ \end{array}$ $\begin{array}{rlr}& \swarrow & \searrow \\ \\ 3x-2& =0& \text{or}{\textstyle \phantom{\rule{1em}{0ex}}}x+2& =0& {\text{Zero product property}}\\ \\ x& ={\displaystyle \frac{2}{3}}& \text{or}{\textstyle \phantom{\rule{2em}{0ex}}}x& =-2\end{array}$

The $x$ -intercepts of the graph of $y=f(x)$ are $({\displaystyle \frac{2}{3}},0)$ and $(-2,0)$ .

Our work also shows that $\frac{2}{3}$ is a zero of multiplicity $1$ and $-2$ is a zero of multiplicity $2$ . This means that the graph will cross the $x$ -axis at $({\displaystyle \frac{2}{3}},0)$ and touch the $x$ -axis at $(-2,0)$ .

### Finding the end behavior

To find the end behavior of a function, we can examine the leading term when the function is written in standard form.

Let's write the equation in standard form.

The leading term of the polynomial is ${3{x}^{3}}$ , and so the end behavior of function $f$ will be the same as the end behavior of $3{x}^{3}$ .

Since the degree is odd and the leading coefficient is positive, the end behavior will be: as $x\to +\mathrm{\infty}$ , $f(x)\to +\mathrm{\infty}$ and as $x\to -\mathrm{\infty}$ , $f(x)\to -\mathrm{\infty}$ .

### Sketching a graph

We can use what we've found above to sketch a graph of $y=f(x)$ .

Let's start with end behavior:

- As
,$x\to +\mathrm{\infty}$ .$f(x)\to +\mathrm{\infty}$ - As
,$x\to -\mathrm{\infty}$ .$f(x)\to -\mathrm{\infty}$

This means that in the "ends," the graph will look like the graph of $y={x}^{3}$ .

Now we can add what we know about the $x$ -intercepts:

- The graph touches the
-axis at$x$ , since$(-2,0)$ is a zero of even multiplicity.$-2$ - The graph crosses the
-axis at$x$ , since$({\displaystyle \frac{2}{3}},0)$ is a zero of odd multiplicity.$\frac{2}{3}$

Finally, let's finish this process by plotting the $y$ -intercept $(0,-8)$ and filling in the gaps with a smooth, continuous curve.

While we don't know exactly where the turning points are, we still have a good idea of the overall shape of the function's graph!

### Positive and negative intervals

Now that we have a sketch of $f$ 's graph, it is easy to determine the intervals for which $f$ is positive, and those for which it is negative.

We see that $f$ is positive when $x>{\displaystyle \frac{2}{3}}$ and negative when $x<-2$ or $-2<x<{\displaystyle \frac{2}{3}}$ .

## Check your understanding

**1)**You will now work towards a sketch of

## Want to join the conversation?

- How do you match a polynomial function to a graph without being able to use a graphing calculator?(18 votes)
- Seeing and being able to graph a polynomial is an important skill to help develop your intuition of the general behavior of polynomial function. In practice, we rarely graph them since we can tell
**a lot**about what the graph of a polynomial function will look like just by looking at the polynomial itself.

For example, given ax² + bx + c

If a is positive, the graph will be like a U and have a minimum value.

If a is negative, the graph will be flipped and have a maximum value

If |a| is > 1 the parabola will be very narrow.

If |a| is < 1 the parabola will be very wide.

The axis of symmetry (and the location of the vertex) is given by -b/2a.

When you factor the function, you get the x-intercepts (if any).

When you set x=0 you get the y intercept of the graph (if any).

Check out these pages for more information:

https://www.mathsisfun.com/algebra/quadratic-equation-graphing.html

Here is a neat tool where you can change the values of the coefficients of a polynomial and see how it affects the resulting graph - a really great way to develop your "mind's eye" and visualize the graph just by looking at the polynomial.

https://www.mathsisfun.com/algebra/quadratic-equation-graph.html(55 votes)

- What is multiplicity of a root and how do I figure out?(11 votes)
- When you have a factor that appears more than once, you can raise that factor to the number power at which it appears. For example if you have (x-4)(x+3)(x-4)(x+1)

the multiplicities of each factor would be:

x-4= multiplicity 2

x+3+ multiplicity 1

x+1= Multiplicity 1(8 votes)

- Can there be any easier explanation of the end behavior please. The way that it was explained in the text, made me get a little confused. Also, for the practice problem, when ever x equals zero, does it mean that we only solve the remaining numbers that are not zeros?(6 votes)
- This video gives a good explanation of how to find the end behavior:

https://www.khanacademy.org/math/algebra2/polynomial-functions/polynomial-end-behavior/v/polynomial-end-behavior(7 votes)

- How can you graph f(x)=x^2 + 2x - 5? step by step?

(I'm new here)(5 votes)- Given a polynomial in that form, the best way to graph it by hand is to use a table.

My personal rule is to always use the values most easiest, so let's use -2,-1,0,1,2

F(-2) = (-2)²-2(-2)-5 = 4+4-5 = 3 ==> (-2,3)

F(-1) = (-1)²-2(-1)-5 = 1+2-5 = -2 ==> (-1,-2)

F(0) = (0)² -2(0) -5 = 0-0-5 = -5 ==> (0,-5)

F(1) = (1)² -2(1) -5 = 1-2-5 ==> (1,-6)

F(2) = (2)² -2(2)-5 = 4-4-5 = (2,-5)

^ just a reminder that the number inside the f(x) (e.g f(-1)) is the number that you plug in the equation.

As you notice, the value inside the f(x) is the x-coordinate, and the output of that is the y coordinate

(e.g (-1,-2) at f(-1)).

hopefully that helps ! and if you have any suspicious, 100% recommend using desmos to check your work.(4 votes)

- What throws me off here is the way you gentlemen graphed the Y intercept. Surely there is a reason behind it but for me it is quite unclear why the scale of the y intercept (0,-8) would be the same as (2/3,0). Would appreciate an answer. Have a good day!(2 votes)
- I see what you mean, but keep in mind that although the scale used on the X-axis is almost always the same as the scale used on the Y-axis, they do not HAVE TO BE the same.

I can't answer for the designers of the graph, but perhaps they wanted to save space, or perhaps it was done in error --- who knows!(7 votes)

- How do you find the end behavior of your graph by just looking at the equation. how do you determine if it is to be flipped?(3 votes)
- by looking at the highest degree term.

kx^n

if n is even, the two "ends" will be the same. if it's odd they will be different.

k determines what infinity is.

if k is positive, infinity will be positive.

if k is negative, infinity will be negative.

combine the two to figure out negative infinity.(4 votes)

- Question number 2--'which of the following could be a graph for y = (2-x)(x+1)^2' confuses me slightly. Shouldn't the y-intercept be -2?

y = (2-0)(0+1)(0+1)

y = (-2)(1)(1)

y = -2

What makes the y-intercept positive 2 instead of negative 2?(3 votes)- You have a math error. 2-0 = 2, not -2.

So, the y-intercept is at y=+2(3 votes)

- You should put a practice for this topic(4 votes)
- i cant understand the second question 2) Which of the following could be the graph of y=(2-x)(x+1)^2y=(2−x)(x+1)

2(3 votes)- Well you could start by looking at the possible zeros. Since the factors are (2-x), (x+1), and (x+1) (because it's squared) then there are two zeros, one at x=2, and the other at x=-1 (because these values make 2-x and x+1 equal to zero).

Since (x+1) is squared, it has multiplicity 2, which means there's two of them in the factor list. This results in the line of the graph just barely touching zero, rather than crossing it. So you're looking for a graph with zeros at x=-1 and x=2, crossing zero only at x=2.

Then you determine the end behavior by multiplying all the factors out using algebra, and it has a negative leading coefficient and an odd exponent, which means the end behavior will be x -> (inf) y -> (-inf), and x -> (-inf) y -> (inf).

Hope this helps!!(2 votes)

- How do you find the end behavior of your graph by just looking at the equation. how do you determine if it is to be flipped?(3 votes)